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If $\displaystyle A = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots\cdots\cdots+\frac{1}{\sqrt{999}}+\frac{1}{\sqrt{1000}}.$

Then $\lfloor A \rfloor$ is, where $\lfloor A\rfloor = A-\{A\}.$

$\bf{My\; Try::}$ Using $$\left(\sqrt{k}+\sqrt{k-1}\right)<2\sqrt{k}<\left(\sqrt{k+1}+\sqrt{k}\right)\;,$$ where $k\in N$ and $k\geq 2$

Then $$\displaystyle \left(\sqrt{k+1}-\sqrt{k}\right)<\frac{1}{2\sqrt{k}}<\left(\sqrt{k}-\sqrt{k-1}\right)$$

So $$\displaystyle 2\sum_{k=2}^{1000}\left(\sqrt{k+1}-\sqrt{k}\right)<\sum_{k=2}^{1000}\frac{1}{\sqrt{k}}<2\sum_{k=2}^{1000}\left(\sqrt{k}-\sqrt{k-1}\right)$$

So we get $$\displaystyle 2(10\sqrt{10}-\sqrt{2}) < A < 2(10\sqrt{10}-1)$$

So we get $$60.41\approx <A<\approx 61.24$$. So $$\lfloor A \rfloor = 61$$

My Question is how can we solve using Concept of definite Integral, plz explain me

Thanks

Community
  • 1
juantheron
  • 53,015

2 Answers2

2

Notice that:

$$60.41<60.5<61.24$$

but,

$$\lfloor60.5\rfloor=60\ne61$$

so your logic is wrong. To correct it, simply note that:

$$\sum_{k=2}^{1000}\frac1{\sqrt k}=\sum_{a=2}^{10}\frac1{\sqrt a}+\sum_{b=11}^{1000}\frac1{\sqrt b}$$

Likewise, you will find that

$$\sum_{a=2}^{10}\frac1{\sqrt a}\approx4.02100$$

$$56.64392\approx2(\sqrt{1001}-\sqrt{11})<\sum_{b=11}^{1000}\frac1{\sqrt b}<2(\sqrt{1000}-\sqrt{10})\approx56.92100$$

Thus,

$$60.66492<\sum_{k=2}^{1000}\frac1{\sqrt k}<60.94200$$

And so we see the correct answer is actually $60$.

Another approach would to brute force it with the Euler-Maclaurin formula:

$$\sum_{k=1}^n\frac1{\sqrt k}=\zeta(1/2)+2\sqrt n+\frac1{2\sqrt n}+\mathcal O(n^{-3/2})$$

where $\zeta(1/2)\approx-1.4603545088$, and this gives us

$$\sum_{k=2}^{1000}\frac1{\sqrt k}\approx-1+\zeta(1/2)+2\sqrt{1000}+\frac1{2\sqrt{1000}}=60.80101$$

with an error at most of $\frac1{24\times1000^{3/2}}$, so we're pretty safe.

Simply Beautiful Art
  • 74,685
2

With $k$ integer and $$x< k< x+1$$

we have

$$\frac1{\sqrt{x+1}}<\frac1{\sqrt k}<\frac1{\sqrt x}.$$

Then integrating in unit intervals,

$$\int_1^{1000}\frac{dx}{\sqrt{x+1}}<\sum_{k=2}^{1000}\frac1{\sqrt k}<\int_1^{1000}\frac{dx}{\sqrt x},$$ i.e.

$$60.449\approx2(\sqrt{1001}-\sqrt2)<\sum_{k=2}^{1000}\frac1{\sqrt k}<2(\sqrt{1000}-\sqrt1)\approx61.245$$

Unfortunately, this bracketing does not allow us to conclude, as the floor could be $60$ or $61$. We can tighten by pulling the first terms out,

$$\color{green}{60}.520\approx2(\sqrt{1001}-\sqrt3)+\frac1{\sqrt2}<\sum_{k=2}^{1000}\frac1{\sqrt k}<2(\sqrt{1000}-\sqrt2)+\frac1{\sqrt2}\approx\color{green}{61}.124$$

$$\color{green}{60}.561\approx2(\sqrt{1001}-\sqrt4)+\frac1{\sqrt2}+\frac1{\sqrt3}<\sum_{k=2}^{1000}\frac1{\sqrt k}<2(\sqrt{1000}-\sqrt3)+\frac1{\sqrt2}+\frac1{\sqrt3}\approx\color{green}{61}.066$$

$$\color{green}{60}.589\approx2(\sqrt{1001}-\sqrt5)+\frac1{\sqrt2}+\frac1{\sqrt3}+\frac1{\sqrt4}<\sum_{k=2}^{1000}\frac1{\sqrt k}<2(\sqrt{1000}-\sqrt4)+\frac1{\sqrt2}+\frac1{\sqrt3}+\frac1{\sqrt4}\approx\color{green}{61}.030$$

$$\cdots$$

Further terms confirm $60$.